Physics - Geometrical Optics Question with Solution | TestHub

Snell's law,

$\sin {45}^{\mathrm{o}}=\mathrm{\mu }\sin \mathrm{\alpha }$

$\sin \mathrm{\alpha }=\frac{1}{\sqrt{2}\mathrm{ }\mathrm{\mu }}$

$\cos \alpha =\frac{\sqrt{2{\mu }^2-1}}{\sqrt{2}\mu }$    ___(1)

From $∆\mathrm{A}\mathrm{B}\mathrm{C},\mathrm{ }\mathrm{A}\mathrm{C}=\sqrt{2}\mathrm{ }\mathrm{t}$

From $∆\mathrm{A}\mathrm{D}\mathrm{C},\cos \mathrm{\alpha }=\frac{\mathrm{D}\mathrm{C}}{\mathrm{A}\mathrm{C}}=\frac{{\mathrm{t}}^{\mathrm{\text{'}}}}{\sqrt{2}\mathrm{}\mathrm{t}}$    ___(2)

From (1) and (2)

$\frac{\sqrt{2\mathrm{ }{\mathrm{\mu }}^2-1}}{\sqrt{2}\mu }=\frac{{\mathrm{t}}^{\mathrm{\text{'}}}}{\sqrt{2}\mathrm{ }\mathrm{t}}\Rightarrow t^{\text{'}}=\frac{\left(\sqrt{2\mathrm{ }{\mathrm{\mu }}^2-1}\right)\mathrm{t}}{\mathrm{\mu }}$